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If $DEF$ is the orthic triangle of $\triangle ABC$, then prove that
$$\frac{\text{Perimeter of }\triangle DEF}{\text{Perimeter of }\triangle ABC} = \frac{r}{R} $$ where $r$ and $R$ are the inradius and the circumradius of $\triangle ABC$.

My attempt is very simple , I put the side length of orthic triangle in terms of $\cos$ and the side length of $\triangle ABC$ but I can't get the required answer.

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    $\begingroup$ You could use the following result : The ratio of a side say AB of a triangle to the corresponding side A'B' of the orthic (=pedal) triangle is equal to the ratio of the circumradius $R$ to the distance from the circumcenter to side AB demonstrations.wolfram.com/… $\endgroup$ – Jean Marie Feb 3 '19 at 23:50
  • $\begingroup$ Important point of vocabulary : you shouldn't use here the term "pedal triangle" but "orthic triangle". let me quote from the excellent site "Cut-the-knot" the following sentence : cut-the-knot.org/triangle/PedalTriangle.shtml "Pedal triangle of point P with respect to a given ΔABC is formed by the feet of the perpendiculars from P to the sides of ΔABC. The most popular of these is, probably, the orthic triangle obtained when P=H, the orthocenter of ΔABC". $\endgroup$ – Jean Marie Feb 4 '19 at 8:11
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Let $AD$, $BE$ and $CF$ be altitudes of an acute-angled $\Delta ABC$.

Thus, since $BCEF$ is cyclic, we obtain that $\Delta AEF\sim\Delta ABC,$ which gives $$\frac{FE}{BC}=\frac{AF}{AC}$$ or in the standard notation $$\frac{FE}{a}=\cos\alpha.$$ Id est, $$\frac{P_{\Delta DEF}}{P_{\Delta ABC}}=\frac{\sum\limits_{cyc}a\cos\alpha}{a+b+c}=\frac{\sum\limits_{cyc}\frac{a(b^2+c^2-a^2)}{2bc}}{a+b+c}=\frac{\sum\limits_{cyc}a^2(b^2+c^2-a^2)}{2abc(a+b+c)}=$$ $$=\frac{\sum\limits_{cyc}(2a^2b^2-a^4)}{2abc(a+b+c)}=\frac{16S^2}{2abc(a+b+c)}=\frac{\frac{2S}{a+b+c}}{\frac{abc}{4S}}=\frac{r}{R}.$$ I used the cyclic sum: $$\sum_{cyc}a^2(b^2+c^2-a^2)=a^2(b^2+c^2-a^2)+b^2(c^2+a^2-b^2)+c^2(a^2+b^2-c^2)=$$ $$=2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4=\sum_{cyc}(2a^2b^2-a^4).$$ Also, $$S_{\Delta ABC}=\sqrt{p(p-a)(p-b)(p-c)}=$$ $$=\sqrt{\frac{a+b+c}{2}\cdot\frac{a+b-c}{2}\cdot\frac{a-b+c}{2}\cdot\frac{-a+b+c}{2}}=$$ $$=\sqrt{\frac{1}{16}(2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4)}.$$

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  • $\begingroup$ Rosenberg Can you explain last 3rd and 4th step. How you write 2 a^2b^2 in last 4th step and in last 3rd step how you write 16 S^2 also also explain me how you solve the summation Series. Plz , I can't understand these two steps. $\endgroup$ – saket kumar Feb 4 '19 at 3:55
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    $\begingroup$ That is a fancy way to write that is! $\endgroup$ – Anubhab Ghosal Feb 4 '19 at 9:51
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$$\sin{2A}+\sin{2B}+\sin2C=2\sin{(A+B)}\cos{(A-B)}+2\sin{C}\cos{C}$$$$=2\sin{C}(\cos{(A-B)}-\cos{(A+B)})=4\sin{A}\sin{B}\sin{C}$$

$$\text{Now, }\frac{EF}{\sin A}=\frac{AE}{\sin C}=\frac{c\cos A}{\sin C}\implies EF=2R\sin A\cos A=R\sin{2A}\text{ .}$$

$$\therefore \text{Perimeter of }\triangle DEF=4R\sin{A}\sin{B}\sin{C}=\frac{2\Delta}{R}=\frac{r\cdot 2s}{R}$$$$\implies\frac{\text{Perimeter of }\triangle DEF}{\text{Perimeter of }\triangle ABC} = \frac{r}{R}$$

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Explanation: $\Delta=\frac{1}{2}ab\sin C=\frac{1}{2}(2R\sin A)(2R \sin B)\sin C=2R^2\sin{A}\sin{B}\sin{C}$

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